For f(x) = floor(x), does lim_{x->1} f(x) exist?

For f(x) = floor(x), does lim_{x->1} f(x) exist?

• A. Yes; the limit exists and equals 1.
• B. Yes; the limit exists and equals 0.
• C. No; the left-hand limit is 0 and the right-hand limit is 1, so the limit does not exist.
• D. No; both one-sided limits equal 0.

Answer: (C)

Think about how floor behaves near an integer. For numbers just below 1, floor(x) is 0, while for numbers just above 1, floor(x) is 1. So the left-hand limit as x approaches 1 is 0 and the right-hand limit is 1. Since these one-sided limits don’t agree, the two-sided limit does not exist. The function value at x = 1 is 1, but the limit concerns values as x approaches 1 from both sides, not the value at the point. This jump at integers explains why the limit fails to exist.

Think about how floor behaves near an integer. For numbers just below 1, floor(x) is 0, while for numbers just above 1, floor(x) is 1. So the left-hand limit as x approaches 1 is 0 and the right-hand limit is 1. Since these one-sided limits don’t agree, the two-sided limit does not exist. The function value at x = 1 is 1, but the limit concerns values as x approaches 1 from both sides, not the value at the point. This jump at integers explains why the limit fails to exist.